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multiphase interfoam k-omega BC

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Old   October 7, 2020, 22:58
Default multiphase interfoam k-omega BC
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Diego Villegas Díaz
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Hi, I am a new foamer. I am simulating a section of an openchanel steady water flow with interFoam solver.

The inlet condition is a water profile and a constant flow in. To represent this situation I used this boundary condition:

1) alpha.water: I used funkySetBoundaryFields to map the waterprofile to alpha.water variable. 1 for water and 0 for air, without intermediate values.

2) U: I used the variableHeightFlowRateInletVelocity with the inlet flow.

3) p_rgh: zero gradient.

4) nut: calulated.

5) k: turbulentIntensityKineticEnergyInlet; with Intensity = 0.0573

6) omega: turbulentMixingLengthFrequencyInlet; with mixing length = 0.00258

I've had some troubles with the k and omega boundary conditions. interFoam fall at the start and I dont understand why. Meanwhile, I am using a fixed value for omega (3.46) or fixed value for k (0.0044065), and its work but I really want to understand why when I use the boundary condition 5) and 6) together interFoam fall.

Thaks in advance.
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Old   October 8, 2020, 01:08
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Charles
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I'm curious how do you get these values, i.e. intensity 0.0573, mixing length 000258, omega 3.46 and k 0.0044065. I guess the divergence was caused by improper initial conditions because turbulence model is prone to diverge.
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Old   October 8, 2020, 08:53
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Diego Villegas Díaz
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Quote:
Originally Posted by Marpole View Post
I'm curious how do you get these values, i.e. intensity 0.0573, mixing length 000258, omega 3.46 and k 0.0044065. I guess the divergence was caused by improper initial conditions because turbulence model is prone to diverge.
I get these values using the water profile at the inlet and a median velocity (U) . I calculated the intensity using some expresion in the book
"Turbulence in open channel flows" (Nezu & Nakagawa) the mixing length scale, k = (3/2)*(I*U)^2, the mixing length for the omega condition I calculated as 0.07*Rh (with Rh the hydraulic radius) I found this scale in these forum but I dont have any formal information. Finally to calculate omega (for fixed value), I calculated another length scale to mixing length lm equal to 0.41*Rh (based in Von Karman constant) and omega is equal to sqrt(k)/lm.

I forgot to mention that my system is initialized at the inlet as uniforms values for k, omega and lm, the same values that I mentioned above. May be these unifor initial condition is the reason why interFoam falls.
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Old   October 8, 2020, 20:10
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k = (3/2)*(I*U)^2 is not mixing length, but is turbulent kinetic energy. So, I guess you need to review these initial values.
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Old   October 8, 2020, 22:41
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Quote:
Originally Posted by Marpole View Post
k = (3/2)*(I*U)^2 is not mixing length, but is turbulent kinetic energy. So, I guess you need to review these initial values.
Ohh I made a mistake writing the last post. I know that k is turbulent Kinect energy. But I'm thinking that interFoam fall due the uniform initial condition un k and omega.
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Old   October 9, 2020, 01:44
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You have two mixing length. The first mixing length 0.07Rh might be problematic and is too small. You can try 0.41Rh instead.
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